DATA606 - Homework 7

7.24 Nutrition at Starbucks, Part I.

The scatterplot below shows the relationship between the number of calories and amount of carbohydrates (in grams) Starbucks food menu items contain. Since Starbucks only lists the number of calories on the display items, we are interested in predicting the amount of carbs a menu item has based on its calorie content.

7.24

1. Describe the relationship between number of calories and amount of carbohydrates (in grams) that Starbucks food menu items contain.

From the scatterplot, there seems to be a positive, moderately strong linear association between calories and carbs.

2. In this scenario, what are the explanatory and response variables?

Explanatory: number of calories Response: amount of carbohydrates (in grams)

3. Why might we want to fit a regression line to these data?

We can predict carbs for a given number of calories using a regression line. This may be useful information for determining whether or not a particular food from Starbucks fits into your diet.

4. Do these data meet the conditions required for fitting a least squares line?

Linearity:The relationship between calories and carbs in the scatterplot appears to be linear.

Nearly Normal Residuals:The residuals is nearly normal as shown in the histogram.

Constant Variability:The variability of residuals around the 0 line has a fan shape on the residuals plot. Thus the data doesn’t meet the conditions required for fitting a least squares line.

7.26 Body measurements, Part III.

Exercise 7.15 introduces data on shoulder girth and height of a group of individuals. The mean shoulder girth is 107.20 cm with a standard deviation of 10.37 cm. The mean height is 171.14 cm with a standard deviation of 9.41 cm. The correlation between height and shoulder girth is 0.67.

1. Write the equation of the regression line for predicting height.

\(\bar{x}=107.20\) \(s_x=10.37\) \(\bar{y}=171.14\) \(s_y=9.41\) correlation \(R=0.67\)

Slope: \(b_1 = \frac{s_y}{s_x}R = \frac{9.41}{10.37}\times0.67 = 0.61\)

Intercept: \(b_0 = \bar{y}-b_1\bar{x} = 171.14-0.61\times107.2 = 105.75\)

Equation of the regression line: \(\hat{y} = \beta_0 + \beta_1 x\) \[height = 105.75+0.61\times shoulder girth\]

2. Interpret the slope and the intercept in this context.

Slope \(b_1\): For each addition cm in should girth, the model predicts an additional 0.61cm in height.

Intercept \(b_0\): When the should girth is 0cm, the height is exected to be 105.75cm. It doesn’t make sense to have a shoulder girth of 0cm in this context. Here, the y-intercept serves only to adjust the height of the line and is meaningless by itself.

3. Calculate R2 of the regression line for predicting height from shoulder girth, and interpret it in the context of the application.

\(R^2 = 0.67^2 = 0.45\) About 45% of the variability in height is accounted for by the model, i.e. explained by the shoulder girth.

4. A randomly selected student from your class has a shoulder girth of 100 cm. Predict the height of this student using the model.

\(height = 105.75+0.61\times shoulder girth\) \(height = 105.75+0.61\times 100 = 166.75cm\)

5. The student from part (d) is 160 cm tall. Calculate the residual, and explain what this residual means.

\(e_i = y_i - \hat{y_i} = 160 - 166.75 = -6.75\) A negative residual means that the model overestimates the height.

6. A one year old has a shoulder girth of 56 cm. Would it be appropriate to use this linear model to predict the height of this child?

No, this calculation would require extrapolation.

7.30 Cats, Part I.

The following regression output is for predicting the heart weight (in g) of cats from their body weight (in kg). The coefficients are estimated using a dataset of 144 domestic cats.

7.30.1

7.30.2

1. Write out the linear model.

\(\hat{heartWeight} = -0.357 + 4.034 \times \hat{bodyWeight}\)

2. Interpret the intercept.

Expected heart weight of a cat with no body weight is -0.357. This is obviously not a meaningful value, it just serves to adjust the height of the regression line.

3. Interpret the slope.

For each additional kilogram increase in body weight of a cat, the model predicts an addition gram in heart weight of the cat.

4. Interpret R2.

The body weight explains 64.66% of the variability in heart weights of cats.

5. Calculate the correlation coefficient.

\(R = \sqrt{R^2} = \sqrt{0.6466} = 0.804\)

7.40 Rate my professor.

Many college courses conclude by giving students the opportunity to evaluate the course and the instructor anonymously. However, the use of these student evaluations as an indicator of course quality and teaching effectiveness is often criticized because these measures may reflect the influence of non-teaching related characteristics, such as the physical appearance of the instructor. Researchers at University of Texas, Austin collected data on teaching evaluation score (higher score means better) and standardized beauty score (a score of 0 means average, negative score means below average, and a positive score means above average) for a sample of 463 professors. The scatterplot below shows the relationship between these variables, and also provided is a regression output for predicting teaching evaluation score from beauty score.

7.40.1

1. Given that the average standardized beauty score is -0.0883 and average teaching evaluation score is 3.9983, calculate the slope. Alternatively, the slope may be computed using just the information provided in the model summary table.

\(b_0 = \bar{y} - b_a\bar{x}\) \(b_1 = \frac{\bar{y} - b_0}{\bar{x}} = \frac{3.9983 - 4.010}{-0.0883} = 0.1325\)

2. Do these data provide convincing evidence that the slope of the relationship between teaching evaluation and beauty is positive? Explain your reasoning.

Since the slope is positive, the relationship between teaching evaluation and beauty is positive. However, they are not strongly correlated.

3. List the conditions required for linear regression and check if each one is satisfied for this model based on the following diagnostic plots.

7.40.2

Linearity: Check using a residuals plot, the relationship between teaching evaluation and beauty appears to be linear.

Nearly normal residuals: Check using a histogram and normal Q-Q plot of residuals, we see that the residuals are nearly normal.

Constant variability: The variability of residuals around the 0 line is roughly constant as well.